Some of the merge operation classes are provided in C++ STL under the header file “algorithm”, which facilitates several merge operations in a easy manner.
Some of them are mentioned below.
- merge(beg1, end1, beg2, end2, beg3) :- This function merges two sorted containers and stores in new container in sorted order (merge sort). It takes 5 arguments, first and last iterator of 1st container, first and last iterator of 2nd container and 1st iterator of resultant container.
-
includes(beg1, end1, beg2, end2) :- This function is used to check whether one sorted container elements are including other sorted container elements or not. Returns true if 1st container includes 2nd container else returns false.
// C++ code to demonstrate the working of // merge() and include() #include<iostream> #include<algorithm> // merge operations #include<vector> // for vector using namespace std;
int main()
{ // Initializing 1st vector
vector< int > v1 = {1, 3, 4, 5, 20, 30};
// Initializing 2nd vector
vector< int > v2 = {1, 5, 6, 7, 25, 30};
// Declaring resultant vector
// for merging
vector< int > v3(12);
// Using merge() to merge vectors v1 and v2
// and storing result in v3
merge(v1.begin(), v1.end(), v2.begin(),
v2.end(), v3.begin());
// Displaying resultant container
cout << "The new container after merging is :\n";
for ( int &x : v3)
cout << x << " ";
cout << endl;
// Initializing new vector
vector< int > v4 = {1, 3, 4, 5, 6, 20, 25, 30};
// Using include() to check if v4 contains v1
includes(v4.begin(), v4.end(), v1.begin(), v1.end())?
cout << "v4 includes v1":
cout << "v4 does'nt include v1";
return 0;
} |
Output
The new container after merging is :
1 1 3 4 5 5 6 7 20 25 30 30
v4 includes v1
Time complexity:
The time complexity of the merge() and include() functions is O(n1 + n2) where n1 and n2 are the sizes of the two containers being merged or checked, respectively.
Space complexity:
The space complexity of the merge() and include() functions is O(n1 + n2) where n1 and n2 are the sizes of the two containers being merged or checked, respectively. This is due to the fact that both functions require an additional container of a size equal to the sum of the sizes of the two containers being merged or checked.
inplace_merge(beg1, beg2, end) :- This function is used to sort two consecutively placed sorted ranges in a single container. It takes 3 arguments, iterator to beginning of 1st sorted range, iterator to beginning of 2nd sorted range, and iterator to last position.
// C++ code to demonstrate the working of // inplace_merge() #include<iostream> #include<algorithm> // merge operations #include<vector> // for vector using namespace std;
int main()
{ // Initializing 1st vector
vector< int > v1 = {1, 3, 4, 5, 20, 30};
// Initializing 2nd vector
vector< int > v2 = {1, 5, 6, 7, 25, 30};
// Declaring resultant vector
// for inplace_merge()
vector< int > v3(12);
// using copy to copy both vectors into
// one container
auto it = copy(v1.begin(), v1.end(), v3.begin());
copy(v2.begin(), v2.end(), it);
// Using inplace_merge() to sort the container
inplace_merge(v3.begin(),it,v3.end());
// Displaying resultant container
cout << "The new container after inplace_merging is :\n";
for ( int &x : v3)
cout << x << " ";
cout << endl;
return 0;
} |
Output:
The new container after inplace_merging is :
1 1 3 4 5 5 6 7 20 25 30 30
set_union(beg1, end1, beg2, end2, beg3) :- This function computes the set union of two containers and stores in new container .It returns the iterator to the last element of resultant container. It takes 5 arguments, first and last iterator of 1st container, first and last iterator of 2nd container and 1st iterator of resultant container . The containers should be sorted and it is necessary that new container is resized to suitable size.
set_intersection(beg1, end1, beg2, end2, beg3) :- This function computes the set intersection of two containers and stores in new container .It returns the iterator to the last element of resultant container. It takes 5 arguments, first and last iterator of 1st container, first and last iterator of 2nd container and 1st iterator of resultant container . The containers should be sorted and it is necessary that new container is resized to suitable size.
One way to implement set-union and set-intersection in sorted ranges can be found here
// C++ code to demonstrate the working of // set_union() and set_intersection() #include<iostream> #include<algorithm> // for merge operations #include<vector> // for vector using namespace std;
int main()
{ // Initializing 1st vector
vector< int > v1 = {1, 3, 4, 5, 20, 30};
// Initializing 2nd vector
vector< int > v2 = {1, 5, 6, 7, 25, 30};
// Declaring resultant vector
// for union
vector< int > v3(10);
// Declaring resultant vector
// for intersection
vector< int > v4(10);
// using set_union() to compute union of 2
// containers v1 and v2 and store result in v3
auto it = set_union(v1.begin(), v1.end(), v2.begin(),
v2.end(), v3.begin());
// using set_intersection() to compute intersection
// of 2 containers v1 and v2 and store result in v4
auto it1 = set_intersection(v1.begin(),v1.end(),
v2.begin(), v2.end(), v4.begin());
// resizing new container
v3.resize(it - v3.begin());
// resizing new container
v4.resize(it1 - v4.begin());
// Displaying set union
cout << "Union of two containers is : ";
for ( int &x : v3)
cout << x << " ";
cout << endl;
// Displaying set intersection
cout << "Intersection of two containers is : ";
for ( int &x : v4)
cout << x << " ";
cout << endl;
return 0;
} |
Output:
Union of two containers is : 1 3 4 5 6 7 20 25 30
Intersection of two containers is : 1 5 30
set_difference(beg1, end1, beg2, end2, beg3) :- This function computes the set difference of two containers and stores in new container .It returns the iterator to the last element of resultant container. It takes 5 arguments, first and last iterator of 1st container, first and last iterator of 2nd container and 1st iterator of resultant container . The containers should be sorted and it is necessary that new container is resized to suitable size.
set_symmetric_difference(beg1, end1, beg2, end2, beg3) :- This function computes the set symmetric difference of two containers and stores in new container .It returns the iterator to the last element of resultant container. It takes 5 arguments, first and last iterator of 1st container, first and last iterator of 2nd container and 1st iterator of resultant container . The containers should be sorted and it is necessary that new container is resized to suitable size.
// C++ code to demonstrate the working of // set_difference() and set_symmetric_difference() #include<iostream> #include<algorithm> // for merge operations #include<vector> // for vector using namespace std;
int main()
{ // Initializing 1st vector
vector< int > v1 = {1, 3, 4, 5, 20, 30};
// Initializing 2nd vector
vector< int > v2 = {1, 5, 6, 7, 25, 30};
// Declaring resultant vector
// for difference
vector< int > v3(10);
// Declaring resultant vector
// for symmetric_difference
vector< int > v4(10);
// using set_difference() to compute difference
// of 2 containers v1 and v2.
auto it = set_difference(v1.begin(), v1.end(),
v2.begin(), v2.end(), v3.begin());
// using set_symmetric_difference() to compute
// symmetric_difference/ of 2 containers
auto it1 = set_symmetric_difference(v1.begin(),
v1.end(), v2.begin(), v2.end(), v4.begin());
// resizing new container
v3.resize(it - v3.begin());
// resizing new container
v4.resize(it1 - v4.begin());
// Displaying set difference
cout << "Difference of two containers is : ";
for ( int &x : v3)
cout << x << " ";
cout << endl;
// Displaying set symmetric_difference
cout << "symmetric_difference of two containers is : ";
for ( int &x : v4)
cout << x << " ";
cout << endl;
return 0;
} |
Output:
Difference of two containers is : 3 4 20
Symmetric difference of two containers is : 3 4 6 7 20 25
The time complexity of set_difference() and set_symmetric_difference() is O(m+n) where m and n are the size of the two input containers.
The space complexity of both the functions is O(m+n) as the resulting vector would take the same size as the combined size of the two input vectors.
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